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Question
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2
Solution
Applying Pythagoras theorem in ΔACE, we obtain
AC2 + CE2 = AE2 ....(i)
Applying Pythagoras theorem in ΔBCD, we get
BC2 + CD2 = BD2 ....(ii)
Using equations (i) and (ii), we get
AC2 + CE2 + BC2 + CD2 = AE2 + BD2 ...(iii)
Applying Pythagoras theorem in ΔCDE, we get
DE2 = CD2 + CE2
Applying Pythagoras theorem in ΔABC, we get
AB2 = AC2 + CB2
Putting these values in equation (iii), we get
DE2 + AB2 = AE2 + BD2.
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